Question

A block $A$ of mass $m$ is in equilibrium after being suspended from the ceiling with the help of a spring of force constant $k$ . The block $B$ of mass $m$ , strikes the block $A$ with a speed $v$ and sticks to it. The value of $v$ for which the spring just attains its natural length is

• A. $\sqrt{\frac{6 m g^{2}}{k}}$
• B. $\sqrt{\frac{3 m g^{2}}{k}}$
• C. $\sqrt{\frac{2 m g^{2}}{k}}$
• D. $\sqrt{\frac{m g^{2}}{2 k}}$

Answer 1

Answer: (A)

Let the velocity of the combined system just after the collision is $v'$ , then using conservation of linear momentum we get
$mv=2mv′$ ,
$\Rightarrow v′=\frac{v}{2}$
Initial elongation of the spring is
$x \, =\frac{m g}{k}$
Now, using the conservation of mechanical energy we get
$\frac{1}{2}2m\left(v '\right)^{2}+\frac{1}{2}kx^{2}=2mgx$
$m\left(\frac{v^{2}}{4}\right)+\frac{m^{2} g^{2}}{2 k}=\frac{2 m^{2} g^{2}}{k}$
$v=\sqrt{\frac{6 m g^{2}}{k}}$